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$\text{In} \Delta PQR,$ $PD \perp QR$ $\text{such that D lies on QR}$ . $\text{If}$ $PQ = a, PR = b, QD = c$ $\text{and}$ $DR=d$ $\text{, prove that }$ $(a + b) (a - b) = (c + d) (c - d).$

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$\text{Given:- PQR is a triangle and}$ $PD\perp QR$

$PQ = a, PR = b, QD = C, DR = d$

To prove:-

$(a + b) (a - b) = (c + d) (c - d)$

$\text{Proof :- In}$ $\Delta PQD$ $\text{use Pythagoras theorem}$

$\\PQ^{2}=QD^{2}+PD^{2}$

$\\a^{2}=c^{2}+PD^{2}\\a^{2}-c^{2}=PD^{2} \; \; \; \; \; \; \; \; ....(1)$

$\text{In }$ $\Delta PRD$ $\text{use Pythagoras theorem }$

$\\PR^{2}=PD^{2}+DR^{2}\\b^{2}=PD^{2}+d^{2}\\b^{2}-d^{2}=PD^{2}\; \; \; \; \; \; \; ....(2)$

equate equation (1) and (2) we get
$\\a^{2} - c^{2} = b^{2} - d^{2}\\a^{2}-b^{2}=c^{2}-d^{2}$

$(a - b) (a + b) = (c - d) (c + d)$ $\left [ \because a^{2} - b^{2} = (a - b) (a + b) \right ]$

Hence proved.

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